There is no such tendency shown by Ct and therefore [H + ] > [OH – ] the solution is acidic and the pH is less than 7. As discussed in the salt hydrolysis of strong base and weak acid. _{a} and K_{b} as K_{h}.K_{b} = K_{w} Let us calculate the Kb value in terms of degree of hydrolysis (h) and the concentration of salt K_{h} = h dos C and

Question 19. The solubility product of Ag_{2}CrO_{cuatro} is 1 x 10 -12 . What is the solubility of Ag_{2}CrO_{4} in 0.01 M AgNO_{step step step three} solution? Answer: Solubility product of Ag_{2}CrO_{4}, K_{sp} = 1 x 10 -2

## Concern twenty-six

Question 20. Write the expression for the solubility product of Ca_{3}(PO_{4})_{2} Answer: Ca_{3}(PO_{4})_{2} (s) \(\rightleftharpoons\) 3Ca 2+ (3s) + 2PO_{4} 3- (2s) The solubility of Ca_{3}(PO_{4})_{2} is, K_{sp} = [Ca 2+ ] 3 . [PO_{4} 3- ] 2 K_{sp} = (3s) 3 . (2s) 2 K_{sp}= (27 s 3 ) . (4s 2 ) K_{sp} = 108s 5 .

## Rain from PbCI

Question 21. A saturated solution, prepared by dissolving CaF_{2}(s) in water, Glendale escort reviews has [Ca 2+ ] = 3.3 x 10 -4 M. What is the K_{sp}of CaF_{2}? Answer: CaF_{dos (s)} \(\rightleftharpoons\) Ca 2+ _{(aq)} + 2F_{–}_{(aq)} [F_{–}] = 2 [Ca 2+ ] = 2 x 33 x 10 -4 M = 6.6 x 10 -4 M = [Ca 2+ ] [F – ] 2 = (3.3 x 10 -4 ) (6.6 x 10 -4 ) 2 = 1.44 x 10 -10

Question 22. K_{sp} of AgCl is 1.8 x 10 -10 . Calculate molar solubility in 1 M AgNO_{3} Answer: AgCl(s) > Ag + (aq) + Cl – (aq) x = solubility of AgCl in 1M AgNO_{3}

[Cl – ] = x K_{sp} = [Ag + ] [Cl – ] 1.8 x 10 -10 = (1) (x) x = 1.8 x 10 -10 M

Question 23. A particular saturated solution of silver chromate Ag_{2}CrO_{4} has [Ag + ] = 5 x 10 -5 and [CrO_{4}] 2- = 4.4 x 10 M. What is the value of for Ag_{2}CrO_{4}? Answer: Ag_{2}CrO_{4} (s) \(\rightleftharpoons\) 2Ag + (aq) + CrO_{4} 2- (aq) K_{sp} = [Ag + ] 2 [CrO_{4} 2- ] = (5 x 10 -5 ) 2 (4.4 x 10 -4 ) = 1.1 x 10 -12

Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO_{3})_{2} and 0.100 L of 0.2 M NaCl are mixed? (PbCI_{2}) = 1.2 x 10 -5 . Answer: When two or more solutions are mixed, the resulting concentrations are different from the original.